Exercise 3.3.1 (${|\mathrm{GL}_n(F) : \mathrm{SL}_n(F)| = q-1}$)

Proof. Consider the map

(If $A\in {\mathrm{GL}}_n(F)$, then $\det <!--\; FUNCTION\; APPLICATION\; -->(A)\ne 0$, so $\det <!--\; FUNCTION\; APPLICATION\; -->(A)\in F^{\times }$.)

Then

• $\phi$ is a homomorphism: If $A,B\in {\mathrm{GL}}_n(F)$, then

  $$\phi (\mathit{𝐴𝐵})=\det <!--\; FUNCTION\; APPLICATION\; -->(\mathit{𝐴𝐵})=\det <!--\; FUNCTION\; APPLICATION\; -->(A)\det <!--\; FUNCTION\; APPLICATION\; -->(B)=\phi (A)\phi (B).$$

• $\phi$ is surjective: If $\lambda \in F^{\times }$, then $F=\det <!--\; FUNCTION\; APPLICATION\; -->(A)$, where $A=\mathrm{diag}(\lambda ,1,1,\dots ,1)\in {\mathrm{GL}}_n(F)$. So

  $$\phi ({\mathrm{GL}}_n(F))=F^{\times }.$$

• $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )={\mathrm{SL}}_n(F)$:

  $$A\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )⟺\phi (A)=1⟺\det <!--\; FUNCTION\; APPLICATION\; -->(A)=1⟺A\in {\mathrm{SL}}_n(F).$$

By the First Isomorphism Theorem,

Since the cardinality of $F$ is $q$, $|{\mathrm{GL}}_n(F)∕{\mathrm{SL}}_n(F)|=q-1$, so

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